MATH00162019sol

Answers for MATH0016 1. Question 1 (a) The general form of the linear second order PDE for a function φ ( x, y ) is Aφ xx + Bφ xy + Cφ yy = D ( x, y ) (i) A = 1 , B = 1 , C = - 2 Therefore B 2 - 4 AC = 9 > 0 and the PDE is hyperbolic. (ii) A = 1 , B = 0 , C = 1 Therefore B 2 - 4 AC = - 4 < 0 and the PDE is elleptic. (iii) A = 1 , B = - 2 , C = 1 Therefore B 2 - 4 AC = 0 and the PDE is parabolic (b) Using the chain rule we have ∂ ∂s = ∂x ∂s ∂ ∂x + ∂y ∂s ∂ ∂y = ∂ ∂x - β ∂ ∂y ∂ ∂t = ∂x ∂t ∂ ∂x + ∂y ∂t ∂ ∂y = ∂ ∂x - α ∂ ∂y so ∂ 2 ∂s∂t = ( ∂ ∂x - β ∂ ∂y )( ∂ ∂x - α ∂ ∂y ) = ∂ 2 ∂x 2 - ( α + β ) ∂ 2 ∂x∂y + αβ ∂ 2 ∂y 2 = ∂ 2 ∂x 2 + ∂ 2 ∂x∂y - 2 ∂ 2 ∂y 2 Therefore α + β = - 1 and αβ = - 2 It follows that α = 1 and β = - 2 The change of variables is then x = s + t, y = 2 s - t and we have the simplified equation ∂ 2 φ ∂s∂t = x (c) Inverting the change of variable in (b) gives s = x + y 3 1

and t = 2 x - y 3 Therefore the PDE becomes φ st = s + t Integrating gives φ = s 2 2 t + t 2 2 s + f ( s ) + g ( t ) where f and g are arbitrary functions. This gives in terms of x and y φ = ( x + y ) 2 18 2 x - y 3 + (2 x - y ) 2 18 x + y 3 + f ( x + y 3 ) + g ( 2 x - y 3 ) or after simplification φ = 3 x ( x + y )(2 x - y ) 54 + f ( x + y 3 ) + g ( 2 x - y 3 ) (d) The conditions give x 3 9 + f ( x 3 ) + g ( 2 x 3 ) = x 3 9 and 1 3 f 0 ( x 3 ) - 1 3 g 0 ( 2 x 3 ) = 0 Integrating the last equation gives f ( x 3 ) - 1 2 g ( 2 x 3 ) = 0 Therefore f = g = 0. It follows that φ = 3 x ( x + y )(2 x - y ) 54 2. Question 2 (a) Use the Lagrange-Euler equation ∂L ∂y - d dx ( ∂L ∂y 0 ) = 0 This yields y 00 - y = e x whose solution is y = Ae x + Be - x + 1 2 xe x 2

The boundary conditions yield A = 1 and B = 0 Therefore the solution is y = e x + 1 2 xe x . (c) We introduce the Lagrange multiplier λ and consider the functional Z π/ 2 0 [ y 0 2 - y 2 - λ ( xy - 2 π K )] dx where K = π 2 + π 3 24 The Lagrange equation yields - 2 y - λx - d dx [2 y 0 ] = 0 or y 00 + y = - λ 2 x whose solution is y = A cos x + B sin x - λ 2 x ( * ) The boundary conditions yields A = 0 and B = λ π 4 Therefore y = λ π 4 sin x - λ 2 x To find the constant λ we use the constraint. This yields Z π/ 2 0 [ x ( λ π 4 sin x - λ 2 x )] dx = K This gives - λ 2 ( π 2 + π 3 24 ) = K Using the definition of K , we obtain λ = - 2 3